The 3.3V Fixed Regulator was burned with 12V input

Question

I am using TI TPS70933DBVR LDO.

Input: 12.6V (3S 18650 Battery)

Output: fixed 3.3V

When I connect the battery to circuit, the LDO is over heating and after 3 second, it will be burned but if I use 4.2V (1S Battery) instead of 12V, everything is ok. 3.3V output working well.

What is the advice about that?

Answer 1

Your load is the problem. With 12 V input, you're dissipating \$(12-3.3)\ \mathrm{V} · 100\ \mathrm{mA} = 870\ \mathrm{mW}\$. For the DBV version, the datasheet lists an approximate Rthj-a of 212.1 °C/W, meaning that when dissipating 870 mW of power, the part will heat up by \$870\ \mathrm{mW} · 212.1\ \mathrm{°C/W} = 184.6\ \mathrm{°C}\$ above ambient. This is well beyond the rated temperature range of the part.

When operating at 4.2 V input, the power dissipation is a much more reasonable 90 mW, which only heats it to about 20 degrees above ambient; warm enough to notice if you put your finger on it, but not hot by any means.

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The solutions here are simple: either

• Use a more efficient switching regulator, which can convert your 12 V to 3.3 V with much lower losses and thus much less heat produced in the first place, at the cost of complexity and increased noise on your output, or
• Use a linear regulator in a better package for heat dissipation. The TPS70933 also comes in a WSON-6 package ("DRV package" in the table below), which has a rated Rthj-a of only 73.1 °C/W (on the standard test board design), so it should remain cool enough to function, though it will still get pretty toasty. You're still throwing away the same amount of energy here; this package is just able to dissipate heat better.
  

For the first option, you can even choose a buck-boost regulator like the TPS6303x to make it possible to use a single cell input and still get 3.3 V out when the cell discharges down to an end voltage below 3.3 V. Or you can use a simple buck converter like the TPS6214x to step down a two- or three-cell input voltage to 3.3 V with about 90% efficiency.

Answer 2

That package has a junction-to-ambient thermal resistance of 212°C/W. Maximum junction temp is 150°C so assuming an ambient temp of 20°C, the maximum power dissipation is 610 mW. With a Vin of 12.6 V, the maximum current that the package can support is 65 mA.

Heatsinking a SOT23 is not realistic. Either use a switching regulator or a pre-regulator.

Answer 3

I agree with what @Hearth said. In addition, I would like to say that if the current values you need are high in the power domain you need and there is a high voltage difference between the input and output voltages of the LDO, the use of LDO is not ideal for you (exactly like the scenario you are in: 12V - 3.3V = 8.7V This may not seem like a high value at first glance, but experienced people will know that this is actually a very high potential difference, especially for such circuits). The basic formula @Hearth gave works for every LDO, you can do the calculation accordingly then compare the results according to datasheets' thermal information sections.

What you need is a buck converter if you don't want to boost the voltage (for example from 12 Volts to 15 Volts or higher). There are many options for this, of course you will need to do a few simple calculations to choose the Buck converter you need.

You mentioned STM32F030, the current this processor draws at the highest frequency with all peripherals active should be around 100 mA. The actual current you need may be much lower than 100 mA, but if you want to guarantee your power domain, I recommend using a Buck converter just in case.

Note: Buck converters also have their own advantages and disadvantages. You should search for details and read at least a few application notes, then you will have enough data and knowledge to choose the right buck converter for you.

Answer 4

What is the advice about that?

Are you concerned about battery consumption? Use a switching regulator. If not, keep using 3S with 12.6V and use a more powerful regulator (you don't need an LDO and ~75% of the battery will be wasted at the regulator).

With 4.2V (1S) what is the lowest battery voltage? You don't have a lot of room @ 100mA, even with the LDO:

Answer 5

The linear voltage stabilizer dissipate heat. Voltage drop multiplayed with current is the power dissipation. Body of stabilizer can dissipate small wattage, heatsink should be used in case then actual dissipation more than that amount.

Answer 6

You did not mention the heat sink you were using. According to "10.1.1 Thermal Protection" in TPS709 150-mA, 30-V, 1-µA IQ Voltage Regulators With Enable:

Any tendency to activate the thermal protection circuit indicates excessive power dissipation or an inadequate heat sink. (...) The TPS709 internal protection circuitry is designed to protect against overload conditions. This circuitry is not intended to replace proper heat sinking.